It follows from (*) that
\begin{align*}
p(t)&=(-t)(-t)^3-1=t^4-1.
\end{align*}

The eigenvalues of $A$ are the roots of the characteristic polynomial $p(t)$.
Solving $t^4-1=0$, we obtain the eigenvalues
\[\pm 1, \pm i,\]
where $i=\sqrt{-1}$.
Note that $t^4-1=(t-1)(t+1)(t-i)(t+i)$.

Final Exam Problems and Solution. (Linear Algebra Math 2568 at the Ohio State University)

This problem is one of the final exam problems of Linear Algebra course at the Ohio State University (Math 2568).